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Exploring Differential Equations An Interactive, Student-Centered Approach

Geoffrey W. Cox, Ph.D.

Section 9.6 Exercises

Exercises 💡 Conceptual Quiz

📝: Abbreviations.

1.

(a) Select all the true statements below.
Select all the true statements below.
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  • An LHCC equation must have constant coefficients.
  • Correct! Constant coefficients are one of the defining features of LHCC equations.
  • An LHCC equation could contain the independent variable, \(x \text{.}\)
  • Incorrect, LHCC equations are linear, meaning they cannot contain non-linear terms like \(y^2 \text{.}\)
  • \(\ds\quad y' + 3y = 0 \) is an LHCC equation.
  • Correct! This is a first-order linear homogeneous differential equation.
  • A non-homogeneous equation has a non-zero free term.
  • Correct! If the free term is not zero, the equation is non-homogeneous.
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(b) Linear or Nonlinear.
    The equation \(y'' + y \cdot y' - 3y = 0 \) is linear.
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  • True.

  • Incorrect. The term \(y \cdot y' \) makes this equation nonlinear because the function \(y \) and its derivative are multiplied together.

  • False.

  • Incorrect. The term \(y \cdot y' \) makes this equation nonlinear because the function \(y \) and its derivative are multiplied together.
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(c) True-or-False.
The hundredth derivative of \(e^{7x} \) is a like-term with \(e^{7x}\text{.}\)
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  • True
  • Correct!
  • False
  • Incorrect.
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(d) Building Solutions.
    If \(y_1\) and \(y_2\) are solutions to a second-order LHCC equation, then
    \begin{equation*} y = 5y_1 - 2y_2 \end{equation*}
    is also a solution.
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  • True.

  • Correct! By the superposition principle, this combination is also a solution.

  • False.

  • Correct! By the superposition principle, this combination is also a solution.
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(e) Identifying Homogeneous Equations.
Which of the following equations is homogeneous?
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  • \(\quad y'' + 3y' + 2y = x \)
  • Incorrect. The right-hand side is \(x \text{,}\) which makes the equation nonhomogeneous.
  • \(\quad y' + e^x = 0 \)
  • Incorrect. The term \(e^x \) on the right-hand side makes the equation nonhomogeneous.
  • \(\quad y'' + 5y' + 7y = 0 \)
  • Correct! The equation is homogeneous because the right-hand side is zero.
  • \(\quad y'' + y' + \sin x = 0 \)
  • Incorrect. The \(\sin x \) term on the right-hand side makes the equation nonhomogeneous.
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(f) Characteristic equation for first-order LHCC.
What is the characteristic equation for \(y' - 5y = 0\text{?}\)
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  • \(\ds\quad r - 5 = 0\)
  • Correct! The characteristic equation is \(r - 5 = 0\text{.}\)
  • \(\ds\quad r + 5 = 0\)
  • Incorrect. Check the sign of the coefficient of \(y\text{.}\)
  • \(\ds\quad r^2 - 5 = 0\)
  • Incorrect. The characteristic equation for a first-order LHCC is linear, not quadratic.
  • \(\ds\quad 5r - 1 = 0\)
  • Incorrect. Make sure to use the correct coefficients from the original equation.
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(g) Identify the first-order LHCC equation.
Which of the following is a first-order LHCC equation?
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  • \(\ds\quad y'' + y' - y = 0\)
  • Incorrect. This is a second-order equation.
  • \(\ds\quad 3y' + 5y = 0\)
  • Correct! This is a first-order linear homogeneous equation with constant coefficients.
  • \(\ds\quad 2y + y' = 3\)
  • Incorrect. This equation is not homogeneous.
  • \(\ds\quad y' + xy = 0\)
  • Incorrect. This is not a constant coefficient equation.
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(h) The Characteristic Equation.
What is the characteristic equation of the differential equation \(y'' - 5y' + 6y = 0\text{?}\)
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  • \(\quad\ds r^2 - 6r + 5 = 0\)
  • Incorrect. Check the coefficients in the original equation.
  • \(\quad\ds r^2 - 5r + 6 = 0\)
  • Correct! The characteristic equation is formed by replacing \(y''\) with \(r^2\text{,}\) \(y'\) with \(r\text{,}\) and \(y\) with 1.
  • \(\quad\ds r^2 - 5r - 6 = 0\)
  • Incorrect. Be careful with the sign of the free term.
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(i) Give the general form.
Give the general form of a second-order LHCC equation if the characteristic equation has the solution: \(r = -1 \pm i\text{.}\)
  • \(y = C_1 e^{-x} + C_2 e^{ix}\)
  • Incorrect. Complex roots require both cosine and sine terms.
  • \(y = (C_1 + C_2 x) e^{-x}\)
  • Incorrect. This form is used for repeated real roots.
  • \(y = C_1 e^{-x} + C_2 e^{x}\)
  • Incorrect. This form is used for distinct real roots.
  • \(y = e^{-x} (C_1 \cos(x) + C_2 \sin(x))\)
  • Correct! This form is used when the roots are complex.
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(j) Give the general form.
Give the general form of a second-order LHCC equation if the characteristic equation has the solutions \(r_1 = 1\) and \(r_2 = -1\text{.}\)
  • \(y = C_1 e^{x} + C_2 e^{-x}\)
  • Correct! This form is used when the characteristic equation has distinct real roots.
  • \(y = C_1 e^{x} + C_2 x e^{x}\)
  • Incorrect. This form is used for repeated real roots.
  • \(y = (C_1 + C_2 x) e^{x}\)
  • Incorrect. This form is also used for repeated real roots.
  • \(y = e^{x} (C_1 \cos(x) + C_2 \sin(x))\)
  • Incorrect. This form is used for complex roots.
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(k) Roots of the characteristic equation.
What are the roots of the characteristic equation \(r^2 - 5r + 6 = 0\text{?}\)
  • \(r = 2\) and \(r = 3\)
  • Correct! The roots are \(r = 2\) and \(r = 3\text{.}\)
  • \(r = -2\) and \(r = -3\)
  • Incorrect. Check the signs of the roots.
  • \(r = 1\) and \(r = 6\)
  • Incorrect. Ensure you solve the quadratic equation correctly.
  • \(r = 5\) and \(r = 1\)
  • Incorrect. Revisit the quadratic formula to solve for the roots.
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(l) General solution for second-order LHCC.
What is the general solution for \(y'' - 5y' + 6y = 0\text{?}\)
  • \(y = C_1 e^{2x} + C_2 e^{3x}\)
  • Correct! The general solution is \(y = C_1 e^{2x} + C_2 e^{3x}\text{.}\)
  • \(y = C_1 e^{-2x} + C_2 e^{-3x}\)
  • Incorrect. Check the signs of the exponents.
  • \(y = C_1 e^{5x} + C_2 e^{x}\)
  • Incorrect. Make sure to use the correct roots.
  • \(y = C_1 e^{x} + C_2 e^{-x}\)
  • Incorrect. Revisit the roots of the characteristic equation.
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(m) Classifying Practice.
Select each classification label that applies to the equation
\begin{equation*} y'' = y' + 6y \end{equation*}
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  • Linear
  • Correct, each of the terms are linear.
  • Homogeneous
  • Correct, the free term is zero.
  • Constant Coefficients
  • Correct, each coefficient is constant.
  • LHCC
  • Correct!
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(n) Classifying Practice.
Select each classification label that applies to the equation
\begin{equation*} 3y''' + y'- \sin(y) = 0 \end{equation*}
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  • Linear
  • Incorrect, \(\sin(y)\) is a nonlinear term.
  • Homogeneous
  • Technically, only linear equations can be labeled as homogeneous or not. Since the equation is nonlinear, we do not select it.
  • Constant Coefficients
  • Technically, only linear equations can be labeled as having constant coefficients or not. Since the equation is nonlinear, we do not select it.
  • LHCC
  • Incorrect.
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(o) Classifying Practice.
Select each classification label that applies to the equation
\begin{equation*} y''- 6 = 0 \end{equation*}
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  • Linear
  • Correct, both terms are linear.
  • Homogeneous
  • Incorrect, the free term, \(6\text{,}\) is non-zero.
  • Constant Coefficients
  • Correct, each coefficient is constant.
  • LHCC
  • Incorrect.
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(p) Classifying Practice.
Select each classification label that applies to the equation
\begin{equation*} \dfrac{d^3y}{dt^3} + k\dfrac{dy}{dt} = ty, \qquad k \text{ is constant} \end{equation*}
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  • Linear
  • Correct, all terms are linear.
  • Homogeneous
  • Correct, the free term is zero.
  • Constant Coefficients
  • Incorrect, the \(y\) term coefficient, \(t\text{,}\) is not constant.
  • LHCC
  • Incorrect.
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(q) Natural Solutions.
Why do exponentials naturally arise as solutions to LHCC equations?
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  • Because their derivatives preserve the same functional form.
  • Correct! Exponentials are eigenfunctions of the derivative operator.
  • Because polynomial solutions cannot satisfy these equations.
  • Incorrect. Polynomial solutions can satisfy LHCC equations.
  • Because sine and cosine functions do not work.
  • Incorrect. Sine and cosine functions are also solutions to LHCC equations.
  • Because they minimize the characteristic equation.
  • Incorrect. Exponentials are not solutions because they minimize the characteristic equation.
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(r) What exponential term is in the solution.
What is the fundamental exponential solution for the equation
\begin{equation*} -2y' - 7y = 0\text{?} \end{equation*}
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  • \(\ds\quad e^{-7/2x}\)
  • Correct! Solving \(-2r - 7 = 0\) gives \(r = -\dfrac{7}{2}\text{.}\) So \(\ds e^{-7/2x}\) is the exponential term in the solution.
  • \(\ds\quad e^{7/2x}\)
  • Incorrect. Check the signs when solving the characteristic equation.
  • \(\ds\quad e^{-2/7x}\)
  • Incorrect. Ensure you are solving the characteristic equation correctly.
  • \(\ds\quad e^{7x}\)
  • Incorrect. Write down the characteristic equation and solve for \(r\text{.}\)
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(s) Structure of the General Solution.
What is the general solution of a second-order LHCC equation if the characteristic equation has unequal real solutions \(r_1\) and \(r_2\text{?}\)
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  • \(\quad y = c_1 e^{r_1 x} + c_1 e^{r_2 x}\)
  • Incorrect. The constants must be different for each term.
  • \(\quad y = c_1 e^{r_1 x} + c_2 e^{r_2 x}\)
  • Correct! Each fundamental solution is multiplied by an arbitrary constant.
  • \(\quad y = e^{r_1 x} + e^{r_2 x}\)
  • Incorrect. The general solution includes arbitrary constants.
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(t) General Solution from Repeated Roots.
Suppose the characteristic equation has a triple root at \(r = -1\text{.}\) What is the corresponding part of the general solution?
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  • \(\quad c_1 e^{-x} + c_2 x e^{-x} + c_3 x^2 e^{-x}\)
  • Correct! Each repeated root contributes an extra power of \(x\text{.}\)
  • \(\quad c_1 e^{-x} + c_2 e^{-2x} + c_3 e^{-3x}\)
  • Incorrect. These are distinct roots, not repeated instances of \(r = -1\text{.}\)
  • \(\quad c_1 e^{-x} + c_2 x^2 e^{-x} + c_3 x^3 e^{-x}\)
  • Incorrect. The powers should stop at \(x^2\) for a triple root.
  • \(\quad c_1 + c_2 x + c_3 x^2\)
  • This form is correct for a root at \(r = 0\text{,}\) not \(r = -1\text{.}\)
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(u) When to Use a Factoring Tool.
Which of the following is the best reason to use a factoring tool or computer algebra system when solving an LHCC equation?
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  • The characteristic equation is high degree and does not factor easily by hand.
  • Correct! Technology saves time and avoids algebraic errors.
  • You already know all the roots from memory.
  • Then you wouldn’t need to factor at all.
  • You want to verify that exponentials are like-terms.
  • That’s a useful idea, but not the purpose of a factoring tool.
  • The roots are all real and distinct.
  • That makes constructing the solution easier—but you still need the roots first.
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(v) Interpreting Mixed Roots.
A characteristic equation has the following roots:
\begin{equation*} r = 0 \ (\text{double}),\quad r = 2 \pm i\text{.} \end{equation*}
What is the correct form of the general solution?
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  • \(\quad y = c_1 + c_2 x + e^{2x}(c_3 \cos x + c_4 \sin x)\)
  • Correct! The double root at 0 gives \(1\) and \(x\text{,}\) and the complex pair gives the oscillatory exponential term.
  • \(\quad y = c_1 + c_2 x + c_3 \cos(2x) + c_4 \sin(2x)\)
  • Incorrect. The sine and cosine terms must be multiplied by \(e^{2x}\text{.}\)
  • \(\quad y = c_1 e^{0x} + c_2 x e^{0x} + c_3 e^{ix} + c_4 e^{-ix}\)
  • While technically valid, this form is not simplified. Use real-valued functions.
  • \(\quad y = c_1 e^{x} + c_2 e^{-x} + c_3 e^{2x}\)
  • These roots do not match the ones given.
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(w) Definition of Homogeneity.
What makes a differential equation homogeneous?
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(x)
Suppose an LHCC equation has \(x^2\) as a solution. Can you conclude anything specific about the structure of such an LHCC equation?
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Exercises 🏋️‍♂️ Practice Drills

1. Identifying LHCC Equations.

(a) Identify the Linear Equations.
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(b) Identify the Linear Homogeneous Equations.
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(c) Identify the Linear Constant Coefficient Equations.
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(d) Identify the LHCC Equations.
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(e) Match the Label to the DE.
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(f) Select the LHCC Equations.
Hint.
There are only 4 LHCC equations in this set.
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(g) Match the DE to Its Characteristic Equation.
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(h) Differential & Characteristic Equation Matching.
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(i)
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(j)
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Differential Equation \(\to\) Characteristic Equation.

Give the characteristic equation corresponding to each differential equation below.
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Use \(r\) for the variable in your answer. Don’t forget the “\(=\)” sign.
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2. \(y'' + 5y' - y = 0\).
\(\displaystyle y'' + 5y' - y = 0\)
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characteristic equation:
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3. \(w''' - 7.1w'' + 0.1w = 0\).
\(\displaystyle w''' - 7.1w'' + 0.1w = 0\)
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characteristic equation:
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4. \(2 p'' + 5p' = 0\).
\(\displaystyle 2 p'' + 5p' = 0\)
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characteristic equation:
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Factor the Characteristic Equation.

Fully factor each characteristic equation.
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⚠️ Warning: The numbers on these problems are randomly generated when you press “Activate”. So, click “Activate” before you start!
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5. \(2\)nd Degree.
\(\displaystyle {r^{2}+r-20}\ =\ \)
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6. \(3\)rd Degree.
\(\displaystyle {r^{3}+5r^{2}-24r}\ =\ \)
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7. \(4\)th Degree.
\(\displaystyle {r^{4}-13r^{2}+36}\ =\ \)
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8. \(3\)rd Degree.
\(\displaystyle {r^{3}-6r^{2}-r+6}\ =\ \)
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Exercises ✍🏻 Problems

From Characteristic Roots \(\to\) General Solution.

Suppose the solutions to a characteristic equation for an LHCC equation are given below. In each case, find the corresponding general solution.
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1.
\(r = 3, -3, 5.3\)
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2.
\(r = 6 \pm i\sqrt{7.7}, 0\)
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3.
\(r = -4, -4, -4, 5.3\)
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4.
\(r = \pm\dfrac{i}{2}, 2 \pm i\)
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5.
\(r = 0, 0, 3, 3, 3, 3\)
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6.
\(r = \pm i, \pi, \pi, 5\)
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7. Verifying Superposition.

Consider the second-order LHCC equation:
\begin{equation} y'' - 3y' + 2y = 0.\tag{9.8} \end{equation}
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Suppose we have already determined that \(y_1 = e^x\) and \(y_2 = e^{2x}\) are solutions. Show that their linear combination
\begin{equation*} y = c_1 e^x + c_2 e^{2x} \end{equation*}
is also a solution for arbitrary constants \(c_1\) and \(c_2\text{.}\)
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Solve the Differential Equations.

Find the general solution for the LNCC equation with the given characteristic equation.
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8.
\(\ds (r-1)^2(r+3)(r^2 + 2r + 5)^2 = 0 \)
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9.
\(\ds (r+1)^2 (r-6)^3(r^2+1)(r^2 + 4) = 0\)
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10.
\(\ds (r-1)^2(r+3)(r^2 + 2r + 5)^2 = 0 \)
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11.
\(\ds (r+1)^2 (r-6)^3(r^2+1)(r^2 + 4) = 0\)
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LHCC Equation General Solution.

Find the general solution to each of the equations below.
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12.
\(\ds y' + \pi y = 0\)
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13.
\(\ds \omega'' - 7\omega' = 0\)
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14.
\(\ds 2\dfrac{d^3 M}{dt^3} - 6\dfrac{dM}{dt} = 0\)
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General Solution.

Find the general solution for each LNCC equation.
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15.
\(4y'' - 36y' = 0\)
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16.
\(4y'' -36y = 0\)
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17.
\(y''- y' - 11y = 0\)
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18.
\(2\dfrac{d^2 \theta}{dt^2} -6\dfrac{d\theta}{dt} - 8\theta = 0\)
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19.
\(m'' = 2m' - m\)
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20.
\(\quad y'' + 4y' + 53y = 0\)
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21.
\(\quad\ds z''=-36z\)
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22.
\(\quad y'' = -24y' - 144y\)
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23.
\(\quad\dfrac{d^2w}{dx^2} - 49w = 0\)
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24.
\(\quad\dfrac{d^2w}{dx^2} + 49w = 0 \)
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25.
\(\quad z''- z' - 42z = 0\)
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26.
\(\quad 9y' + 2y = 0\)
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27.
\(\quad 3y'' + 4y' = 0\)
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28.
\(\quad w'' + 6w' + 9w = 0\)
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29.
\(\quad m'' = 2m' - m\)
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30.
\(\quad y^{(4)} - 5y'' + 4y = 0\)
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Initial-Value Problems.

Solve the following initial value problems.
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31.
\(\quad 4y'' -36y = 0, \ \ y(0) = 4,\ \ y'(0) = -6 \)
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32.
\(\quad\dfrac{d^2z}{dx^2} - 4\dfrac{dz}{dx} + 4z = 0, \ \ z(1) = 1, \ \ z'(1) = 1\)
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33.
\(\quad 2\dfrac{d^2 \theta}{dt^2} -6\dfrac{d\theta}{dt} - 8\theta = 0,\ \ \theta(0) = 12, \ \ \theta'(0) = -2\)
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34. The General Solution to First-Order LHCC Equations.

Derive the general solution to a first order linear homogeneous constant coefficient (LHCC) equation of the form:
\begin{equation*} a_1\ y' + a_0\ y = 0\text{,} \end{equation*}
where \(a_1,\ a_0\) are constants.
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Boundary Value Problems.

Solve the following boundary value problems. Explain whether you found a unique solution, an infinite number of solutions, or no solution.
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35.
Consider the following differential equation:
\begin{equation*} y'' + y = 0, \quad y(0)=0, \quad y\left(b\right)=y_b, \quad b > 0\text{.} \end{equation*}
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Notice that instead of initial conditions at \(t=0\) only, there is one at \(t=b\text{.}\) This is called a boundary-value problem (BVP) because it has conditions over two different \(t\)-values. Now, for this equation find the following, if possible:
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  1. General solution.
    Solution.
    This DE is linear, homogeneous, and has constant coefficients, so we can start by solving the characteristic equation.
    \begin{align*} r^2 + 1 \amp = 0 \\ r^2 \amp = -1 \\ r \amp = 0 \pm i \end{align*}
    Thus, the general solution is
    \begin{align*} y \amp = c_1e^{0t}\sin(t) + c_2e^{0t}\cos(t) \\ \amp = c_1\sin(t) + c_2\cos(t). \end{align*}
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    Answer.
    \(\ds y = c_1\sin(t) + c_2\cos(t) \) (o \(\ds y = c_1\sin(x) + c_2\cos(x) \) )
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  2. The solution with boundary conditions: \(y(0)=0\) and \(y\left(\sfrac{\pi}{2}\right)=0.4\text{.}\)
    Solution.
    We can begin with the general solution we found previously. \(\ds y = c_1\sin(x) + c_2\cos(x). \) When we apply the first condition \(\ds y(0) = 0, \) we have the following.
    \begin{align*} y(0) \amp = 0 \\ c_1\sin(0) + c_2\cos(0) \amp = 0 \\ c1\cdot 0 + c_2 \cdot 1 \amp = 0 \\ c_2 \amp = 0 \end{align*}
    And using the second condition yields the following.
    \begin{align*} y\!\left(\dfrac{\pi}{2}\right) \amp = 0.4 \\ c_1\sin\left(\dfrac{\pi}{2}\right) + c_2\cos\left(\dfrac{\pi}{2}\right) \amp = 0.4 \\ c_1 \cdot 1 + c_2 \cdot 0 \amp = 0.4 \\ c_1 \amp = 0.4 \end{align*}
    Hence, the (unique) particular solution is
    \begin{align*} y \amp = c_1\sin(x) + c_2\cos(x) \\ \amp = 0\cdot\sin(x) + 0.4\cos(x) \\ \amp = 0.4\cos(x) \end{align*}
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    Answer.
    the unique solution is \(\ds y = 0.4\cos(x) \)
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  3. The solution with boundary conditions: \(y(0)=0\) and \(y(pi)=5\text{.}\)
    Solution.
    We can begin with the general solution we found previously. \(\ds y = c_1\sin(x) + c_2\cos(x). \) When we apply the first condition \(\ds y(0) = 0, \) we have the following.
    \begin{align*} y(0) \amp = 0 \\ c_1\sin(0) + c_2\cos(0) \amp = 0 \\ c1\cdot 0 + c_2 \cdot 1 \amp = 0 \\ c_2 \amp = 0 \end{align*}
    And using the second condition yields the following.
    \begin{align*} y(\pi) \amp = 5 \\ c_1\sin(\pi) + c_2\cos(\pi) \amp = 5 \\ c_1 \cdot 0 + c_2 \cdot (-1) \amp = 5 \\ c_2 \amp = -5 \end{align*}
    In this case, the first condition leads t \(\ds c_2 = 0, \) while the second condition leads t \(\ds c_2 = 5. \) These cannot both be true, so there is no solution.
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    Answer.
    no solution
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  4. The solution with boundary conditions: \(y(0)=0\) and \(y(pi)=0\text{.}\)
    Solution.
    We can begin with the general solution we found previously. \(\ds y = c_1\sin(x) + c_2\cos(x). \) When we apply the first condition \(\ds y(0) = 0, \) we have the following.
    \begin{align*} y(0) \amp = 0 \\ c_1\sin(0) + c_2\cos(0) \amp = 0 \\ c1\cdot 0 + c_2 \cdot 1 \amp = 0 \\ c_2 \amp = 0 \end{align*}
    And using the second condition yields the following.
    \begin{align*} y(\pi) \amp = 0 \\ c_1\sin(\pi) + c_2\cos(\pi) \amp = 0 \\ c_1 \cdot 0 + c_2 \cdot (-1) \amp = 0 \\ c_2 \amp = 0 \end{align*}
    In this case, both of the conditions lead to \(\ds c_2 = 0, \) and we don’t learn anything about \(\ds c_1. \) Thus \(\ds c_1 \) could still take on any value, and therefore there are an infinite number of solutions of the form
    \begin{align*} y \amp = c_1\sin(x) + c_2\cos(x) \\ \amp = c_1\sin(x) + 0\cdot\cos(x) \\ \amp = c_1\sin(x). \end{align*}
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    Answer.
    there are an infinite number of solutions of the for \(\ds y = c_1\sin(x) \)
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36.
If possible, solve the following boundary value problem:
\begin{equation*} y'' - y = 0, \quad y(0) = 1, \quad y(1) = 2e- \dfrac{1}{e} \end{equation*}
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37.

Show why \(x\) is needed in the general solution for repeated roots of the CE
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